Question

(a) What volume of 0.115 \(\mathrm{M} \mathrm{HClO}_{4}\) solution is needed to neutralize 50.00 \(\mathrm{mL}\) of 0.0875 \(\mathrm{M} \mathrm{NaOH} ?(\mathbf{b})\) What volume of 25.8 \(\mathrm{mL}\) of an AgNO \(_{3}\) solution is needed to precipitate all the \(\mathrm{Cl}^{-}\) ions in a 785 -mg sample of \(\mathrm{KCl}\) (forming AgCl), what is the molarity of the AgNO \(_{3}\) solution? (d) If 45.3 \(\mathrm{mL}\) of a 0.108 \(M \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{KOH}\) , how many grams of \(\mathrm{KOH}\) must be present in the solution?

Short answer

(a) 38.04 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH. (b) The molarity of AgNO3 solution needed to precipitate all Cl- ions in the 785-mg sample of KCl is 0.408 M. (d) The mass of KOH present in the solution is 0.274 g.

Step-by-step solution

Write a balanced chemical equation

First, we need to write a balanced chemical equation for the neutralization reaction: HClO4(aq) + NaOH(aq) → NaClO4(aq) + H2O(l)

Calculate the moles of NaOH

Next, we need to calculate the moles of NaOH using its given volume and molarity: moles of NaOH = (volume x molarity) moles of NaOH = (50.00 mL) x (0.0875 mol/L) moles of NaOH = \(4.375 \times 10^{-3} mol\)

Calculate the moles of HClO4

Since the stoichiometry is 1:1 between HClO4 and NaOH in the balanced chemical equation, the moles of HClO4 are the same as the moles of NaOH: moles of HClO4 = moles of NaOH moles of HClO4 = \(4.375 \times 10^{-3} mol\)

Calculate the volume of HClO4 solution

Now, we can calculate the volume of the HClO4 solution using the moles of HClO4 and its given molarity: volume of HClO4 = (moles of HClO4) / (molarity of HClO4) volume of HClO4 = \(\frac{4.375 \times 10^{-3} mol}{0.115 mol/L}\) volume of HClO4 = \(38.04 mL\) Therefore, the volume of 0.115M HClO4 solution needed to neutralize 50.00mL of 0.0875M NaOH is 38.04 mL. #(b) Volume and molarity of AgNO3 solution needed to precipitate all Cl- ions in a 785-mg sample of KCl#

Write a balanced chemical equation

First, we need to write a balanced chemical equation for the precipitation of AgCl: AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)

Calculate moles of KCl

We are given the mass of the KCl and can use its molar mass to find the moles of KCl: moles of KCl = (mass of KCl) / (molar mass of KCl) moles of KCl = (785 mg) x (1 g/1000 mg) x \(\frac{1 mol}{74.55 g}\) moles of KCl = \(1.053 \times 10^{-2} mol\)

Calculate moles of AgNO3

In the balanced chemical equation, the stoichiometry between AgNO3 and KCl is 1:1. Therefore, the moles of AgNO3 are the same as the moles of KCl: moles of AgNO3 = moles of KCl moles of AgNO3 = \(1.053 \times 10^{-2} mol\)

Calculate the volume of AgNO3 solution

We are given the volume of AgNO3 solution (25.8 mL), and we can use that to find the molarity of AgNO3: molarity of AgNO3 = (moles of AgNO3) / (volume of AgNO3) molarity of AgNO3 = \(\frac{1.053 \times 10^{-2} mol}{25.8 \times 10^{-3} L}\) molarity of AgNO3 = \(0.408 M\) Therefore, the molarity of AgNO3 solution needed to precipitate all Cl- ions in the 785-mg sample of KCl is 0.408 M. #(d) Mass of KOH present in the solution#

Write a balanced chemical equation

First, we need to write a balanced chemical equation for the neutralization reaction: HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

Calculate moles of HCl

We are given the volume and molarity of HCl, so we can find the moles of HCl: moles of HCl = (volume x molarity) moles of HCl = (45.3 mL) x (0.108 mol/L) moles of HCl = \(4.889 \times 10^{-3} mol\)

Calculate moles of KOH

The stoichiometry between HCl and KOH is 1:1 in the balanced chemical equation. Therefore, the moles of KOH are the same as the moles of HCl: moles of KOH = moles of HCl moles of KOH = \(4.889 \times 10^{-3} mol\)

Calculate the mass of KOH

Now, we can use the moles of KOH and its molar mass to find the mass of KOH: mass of KOH = (moles of KOH) x (molar mass of KOH) mass of KOH = \(4.889 \times 10^{-3} mol\) x \(56.11 \frac{g}{mol}\) mass of KOH = \(0.274 g\) Therefore, the mass of KOH present in the solution is 0.274 g.

Key concepts

Neutralization

Neutralization is a chemical reaction where an acid and a base react to form water and a salt. This process is essential in various chemical applications, including titrations in laboratories. The core principle is that the

• acid provides hydrogen ions (H⁺)
• base provides hydroxide ions (OH^-)

which combine to form water (H₂O). In the example of hydrochloric acid (HCl) neutralizing sodium hydroxide (NaOH), the balanced equation is:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
This equation shows a 1:1 reaction ratio, meaning that one mole of hydrochloric acid reacts with one mole of sodium hydroxide.
Neutralization reactions are useful for determining the concentration of unknown solutions by measuring the volume of a titrant needed to neutralize it.

Precipitation Reactions

Precipitation reactions occur when two soluble salts in aqueous solutions are combined and form an insoluble salt, known as a precipitate. This is a type of double displacement reaction. For example, when silver nitrate (AgNO₃) is mixed with potassium chloride (KCl):
AgNO₃(aq) + KCl(aq) → AgCl(s) + KNO₃(aq)
Here, silver chloride (AgCl) precipitates out of the solution as a solid. The stoichiometry in this balanced equation is 1:1, meaning the moles of silver nitrate required equal the moles of potassium chloride used.
Precipitation reactions are useful for removing ions from solutions, detecting ions in qualitative analysis, and synthesizing chemical compounds. Ensuring the reaction conditions are right, like concentration and temperature, is crucial for maximizing the precipitate formation.

Molar Mass Calculations

Molar mass is a fundamental concept in chemistry representing the mass of one mole of a substance. It's essential for converting between grams and moles, which is crucial in solving stoichiometry problems. To calculate molar mass, sum the atomic masses of all atoms in a compound, based on the periodic table.
For example, in finding the molar mass of potassium chloride (KCl):

• Atomic mass of K: 39.10 g/mol
• Atomic mass of Cl: 35.45 g/mol
• Molar mass of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol

This calculation allows you to convert the mass of KCl into moles, facilitating reactions like determining how much KCl reacts with AgNO₃ to form AgCl in a precipitation reaction. Accurate molar mass calculations are pivotal for precise stoichiometric measurements.

Molarity

Molarity is a way of expressing concentration in chemistry, defined as the number of moles of solute per liter of solution. It is represented with the unit M (mol/L). Calculating molarity is crucial for understanding the exact concentration of solutions involved in reactions like neutralization and precipitation.
To determine molarity:

• Calculate moles of solute (e.g., AgNO₃ or NaOH)
• Measure the volume of the solution in liters
• Use the formula: Molarity (M) = moles of solute / liters of solution

For instance, a problem may provide the amount of solute and the volume of the solution, like in calculating the volume of AgNO₃ needed to react with KCl, allowing the determination of the molarity of the solution.
This concept is vital for preparing solutions of precise concentrations in laboratory settings and in conducting chemical reactions where reactant concentration impacts the outcome.